Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 19 - Planar Kinetics of a Rigid Body: Impulse and Momentum - Section 19.4 - Eccentric Impact - Problems - Page 548: 29


$\omega_2=1.91~ rad/s$

Work Step by Step

We can determine the required angular velocity as follows: According to the conservation of angular momentum $I_{z_1}\omega_1=I_{z_2}\omega_2~~~$[eq(1)] Now $I_{z_1}=mk_z^2$ $\implies I_{z_1}=(0.75)(0.125)^2=0.0117Kg/m^2$ Similarly $I_{z_2}=mk_z^2+\frac{1}{2}mr^2$ We plug in the known values to obtain: $I_{z_2}=(0.75)(0.125)^2+(\frac{0.05}{2})(0.15)^2$ This simplifies to: $I_{z_2}=0.01228Kg/m^2$ We plug in the known values in eq(1) to obtain: $(0.0117)(2)=(0.01228)\omega_2$ This simplifies to: $\omega_2=1.91~ rad/s$
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